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UVA PHYS 632 - Electromagnetic Waves

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Lecture 13 Electromagnetic Waves Chp. 34 Thursday MorningElectromagnetic spectrumEye Sensitivity to ColorProduction of Electromagnetic wavesHow the fields vary at a Point P in space as the wave goes byElectromagnetic WaveProperties of a plane wave of lightA point source of light is one that emits isotropically and the intensity of it falls off as 1/r2The Poynting vector SRadiation pressurePolarization of lightPowerPoint PresentationSlide 14Slide 15Slide 16Reflection and refractionSlide 18Dispersion: Different wavelengths have different indices of refractionSlide 20Slide 21Total Internal reflectionLecture 13 Electromagnetic Waves Chp. 34Thursday Morning•Cartoon -. Opening Demo - Warm-up problem•Physlet •Topics–Light is a Electromagnetic wave–eye sensitivity–Traveling E/M wave - Induced electric and induced magnetic amplitudes–Energy transport Poynting vector–Pressure produced by E/M wave–Polarization –Reflection, refraction,Snell’s Law, Internal reflection–Prisms and chromatic dispersion–Polarization by reflectionElectromagnetic spectrumEye Sensitivity to ColorProduction of Electromagnetic wavesHow the fields vary at a Point P in space as the wave goes byElectromagnetic WaveProperties of a plane wave of light€ E = Emsin(kx −ωt)B = Bmsin(kx −ωt)c =ωk=1μ0ε0= speed of lightλ =2πk€ EmBm= cE and B are always perpendicular to the direction of travelE is perpendicular to BE X B = the direction of travelE and B both vary with the same frequency and in phase.Speed of wave is independent of speed of observerWave doesn’t need a medium to travel in.A point source of light is one that emits isotropically and the intensity of it falls off as 1/r2Let P be the power of the sourcein joules per sec. Then the intensity of light at a distance r is I = P/4r2The Poynting vector S€ S =1μ0E × BS has units of energy/time per unit area or Watt/m2€ S =1μ0EBE is perp. to B and in the energy flows in the direction of the wave. Since B=E/c, we get for the instantaneous power of the waveIntensity I of the wave is defined as Savg € S =1μ0EB =1cμ0E2€ I = Savg=1cμ0Erms2=12cμ0Em2€ Erms=12Em17. The maximum electric field at a distance of 10 m from an isotropic point light source is 2.0 V/m. Calculate(a) the maximum value of the magnetic field and(b) the average intensity of the light there?(c) What is the power of the source?(a) The magnetic field amplitude of the wave is€ Bm=Emc=2.0V m2.998 ×108m s= 6.7 ×10−9T(b) The average intensity is€ Iavg=E2m2μ0c=2.0V m( )22 4π ×10−7T ⋅ m A( )2.998 ×108m s( )= 5.3 ×10−3W m2(c) The power of the source is( )( ) WmWmIrPavg7.6103.510442322=×==−ππRadiation pressure€ Pr=IcThis is the force per unit area felt by an object that absorbs light. (Black piece of paper))€ Pr=2IcThis is the force per unit area felt by an object that reflects light backwards.(Aluminum foil)Polarization of light35. In the figure, initially unpolarized light is sent through three polarizing sheets whose polarizing directions make angles of 1 = 40o, 2 = 20o, and 3 = 40o with the direction of the y axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.)Let Io be the intensity of the unpolarized light that is incident on the first polarizing sheet. The transmitted intensity of is I1 = (1/2)I0, and the direction of polarization of the transmitted light is 1 = 40o counterclockwise from the y axis in the diagram. The polarizing direction of the second sheet is 2 = 20o clockwise from the y axis, so the angle between the direction of polarization that is incident on that sheet and the the polarizing direction of the sheet is 40o + 20o = 60o. The transmitted intensity is,60cos2160cos20212ooIII ==and the direction of polarization of the transmitted light is 20o clockwise from the y axis.35. In the figure, initially unpolarized light is sent through three polarizing sheets whose polarizing directions make angles of 1 = 40o, 2 = 20o, and 3 = 40o with the direction of the y axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.)The polarizing direction of the third sheet is 3 = 40o counterclockwise from the y axis. Consequently, the angle between the direction of polarization of the light incident on that sheet and the polarizing direction of the sheet is 20o + 40o = 60o. The transmitted intensity is.101.360cos2160cos240223−×===ooIIIThus, 3.1% of the light’s initial intensity is transmitted.Reflection and refraction€ θ1= θ'1€ n1sinθ1= n2sinθ2Snells Lawn1n247. In the figure, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. What is the length of the shadow of the pole on the level bottom of the pool?Consider a ray that grazes the top of the pole, as shown in the diagram below. Here 1 = 35o, l1 = 0.50 m, and l2 = 1.50 m. The length of the shadow is x + L. x is given by x = l1tan1 = (0.50m)tan35o = 0.35 m. According to the law of refraction, n2sin2 = n1sin1. We take n1 = 1 and n2 = 1.33 (from Table 34-1). Then,oon55.2533.135sinsinsinsin12112=⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=−−θθL is given by.72.055.25tan)50.1(tan22mmlLo=== θThe length of the shadow is 0.35m + 0.72 m = 1.07 m.21l2l1L xairwatershadowDispersion: Different wavelengths have different indices of refractionTotal Internal reflection€ n1sinθ1= n2sinθ2€ (1.33)sinθ1= (1.00)sin90


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